How can I answer this Putnam question more rigorously?

Why not write it the other way round?

The polynomial function $$F(x)=\sum_{k=0}^n\frac{a_k}{k+1}x^{k+1} $$ is a differentiable function $\Bbb R\to\Bbb R$ with derivative $$F'(x)=\sum_{k=0}^na_kx^k.$$ We are given that $F(1)=0$, and clearly $F(0)=0$. Hence by Rolle's theorem, there exists $x\in(0,1)$ such that $F'(x)=0$, as was to be shown.

Your proof looks fine. If you wanted to expand, you could add the following:

Suppose $f(x)>0$ for all $x>0$. Then we must have $$\int_0^1f(x)\ dx>0,$$ But we have already shown that $\int_0^1f(x)\ dx=0$, a contradiction.

If we assume $f(x)<0$ for all $x>0$, we arrive at a similar contradiction.

You can also prove it using the mean value theorem. You showed that $$\int_{0}^{1}f\left(x\right)dx=0 $$ now since $f$ is continuous by the mean value theorem for integrals we have that exists some $c\in\left(0,1\right) $ such that $$f\left(c\right)=\int_{0}^{1}f\left(x\right)dx $$ so $$f\left(c\right)=0$$ as wanted.